(D) The absorbance of an unknown solution prepared from ground- water was 0.175.
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Question
(D) The absorbance of an unknown solution prepared from ground- water was 0.175. Find the concentration of the unknown. 18-21. Protein concentration. The molar absorptivity of a protein in water at 280 nm can be estimated within ~5-10% from its content of the amino acids tyrosine and tryptophan (Table 10-1) and from the number of disulfide linkages (R-S-S-R) between cysteine residues. 32 280 nm(M-1 cm-1)~ 5 500 nTrp + 1 490 nTy, t 125 ns-s where Trp is the number of tryptophans, nTyr is the number of tyro- sines, and ns-s is the number of disulfides. This information is obtained from the amino acid sequence and from additional studies such as X-ray crystallography and nuclear magnetic resonance to locate disulfide linkages. 457Explanation / Answer
a)
Let us calculate molar absorptivity using given formula
= 8*5500 + 26*1490 + 19*125 = 85115 mol-1 cm-1
b)
The average molecular weight of an amino acid is 110Da. For 678 amino acids it will be
110 * 678 = 74580 Da
Molecular weight of protein is approximately 74580 Da.
Molecular mass of protein with carbohydrates is 79550 Da.
Mass fraction is 74580/79550 = 0.93
0.93/100 = 0.0093 gives us the 1 wt% transferrin.
Now its concentration will be 0.0093/74580 gives us the concentration of transferrin.
c = 1.24 * 10^-7 mol/L
Absorbance of 1wt% of solution will be
A = cl
= 85115 *1.24 * 10^-7 * 1000 = 10.55
c)
Using Beer's Law (A = ebc, where A = absorbance, e=molar absorption coefficient, b is path length, and c is concentration); rearrange to solve for c
c = A/(eb)
= 1.50/(85115*1000) = 1.76 * 10^-8 mol/L
Transferrin molecular weight is 74580 Da = 74580 g/mol
Let us multiply molecular weight with concentration = 1.76 * 10^-8 * 74580 = 0.001 g/L = 0.001 mg/mL
Wt% can be calculated by mass fraction and concentration = 0.93*0.001 g/L= 0.00093
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